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jenyasd209 [6]
2 years ago
11

Upper A 55 ft wire is used to brace a utility pole. If the wire is attached 6 ft from the top of the 40 ft​ pole, how far from t

he base of the pole will the wire be attached to the​ ground
Mathematics
1 answer:
Marysya12 [62]2 years ago
4 0
This seems like it’s to hard!! 2.2.666
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What is the rate of change seen in the graph below?
Levart [38]
(-2,2) eh this not the answer.
3 0
2 years ago
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The area of a sector of a circle is given by the equation , 3.142S/360 where r is the radius of the circle and S is the angle me
Diano4ka-milaya [45]
Given that the area of the sector is given by:
A=3.142r² S/360
where:
r is the radius 
S is the angle measure of the sector
If Mia solved the equation for S, the equations that she wrote would be found as follows;

From A=3.142r² S/360
making S the subject by first dividing both sides by 3.142r^2 we shall have:
A/3.142r²=S/360
next we multiply both sides by 360
360×A/3.142r²=S/360×360
this will give us
(360A)/(3.142r²)=S
therefore the equation that Mia will get is:
S=(360A)/(3.142r²)


4 0
2 years ago
Read 2 more answers
Elaina Kicks A Soccer Ball With An Initial Velocity Of
VladimirAG [237]

By using the motion equations, we conclude that she will make the goal.

<h3>Will she make the goal?</h3>

First, we know that the goal is 20ft away, and the position in x is given by:

x = 28*cos(58)*t

Let's find the time in which the ball will travel these 20 ft.

20 =  28*cos(58)*t

(20)/(28*cos(58)) = t = 1.35

So the horizontal distance is covered in 1.35 seconds, now let's see which is the height of the ball at that time.

The height equation is:

y = -16*t^2 +28*sin(58)*t

Evaluating in t = 1.35 we get:

y = -16*(1.35)^2 +28*sin(58)*1.35\\\\y = 2.9

And the goal is 5ft tall, so we can conclude that she will make the goal.

If you want to learn more about motion equations:

brainly.com/question/19365526

#SPJ1

5 0
1 year ago
(-7)^-1 without an exponent
Verizon [17]
Put the negative exponent as the denominator in a fraction, so you'd get -7/1
3 0
3 years ago
If CDE ~ GDF, find ED
qaws [65]

Answer:

10

Step-by-step explanation:

\triangle CDE \sim \triangle GDF.. (given) \\\\\therefore \frac{CD}{GD} =\frac{DE}{DF}.. (csst) \\\\\therefore  \frac{15}{x+3} =\frac{3x+1}{4}\\\\ \therefore   \: 15 \times 4 = (x + 3)(3x + 1) \\  \\ \therefore   \: 60 = 3 {x}^{2}  + x + 9x + 3 \\  \\ \therefore  3 {x}^{2}  + 10x + 3 - 60 = 0 \\ \therefore  3 {x}^{2}  + 10x  - 57 = 0 \\ \therefore  3 {x}^{2}  + 19x - 9x  - 57 = 0 \\ \therefore   \: x(3x + 19) - 3(3x + 19) = 0 \\\therefore   \:  (3x + 19)(x - 3) = 0 \\ \therefore   \: 3x + 19 = 0 \:  \: or \:  \: x - 3 = 0 \\  \therefore   \: x =  -  \frac{19}{3}  \:  \: or \:  \: x = 3 \\  \because \: x \: can \: not \: be \:  - ve \\ \therefore   \: x = 3 \\ ED = 3x + 1 = 3 \times 3 + 1  \\ \huge \red{ \boxed{ ED= 10}}

7 0
3 years ago
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